Quantum Mechanics I The Fundamentals 1st Rajasekar Solution Manual
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Schrödinger Equation and Wave Function
2.1 Write the Schrödinger equation for a particle in a potential sin x The Schrödinger equation is
2.2 For a particle carrying a charge q in a uniform time-dependent electric field E(t) the force experienced by it is F = ∇( qE(t) ·X). Write the Schrödinger equation of the system. The Schrödinger equation is
2.3 Write the Schrödinger equation for a charged particle moving in an electromagnetic field with the Hamiltonian
Rewrite the Schrödinger equation for A = (Bz, 0,0) and φ = Ez
K24365_SM_Cover.indd 21 13/11/14 6:56 PM 2 2 2
C H A P T E R 2
i ∂ψ = ∂2ψ + sin xψ ∂t 2m ∂x2
i ∂ψ = ∇ ψ (qE(t) · X)ψ . ∂t 2m
1 H = 2m p qA(X,t) 2 c + qφ(X, t)
The Schrödinger equation is 2 i ∂ψ ∂t = 1 p qA 2m c ψ + qφψ 1 2 2 = 2∇ ψ + q A2 ψ p qAψ qA pψ + qφψ 2m c2 c c 1 2 2 = 2∇ ψ + 2i qA ·∇ψ + i q ψ∇· A + q A2ψ + qφψ .
K24365_SM_Cover.indd 21 13/11/14 6:56 PM 2m c c c2 7
8 Solutions to the Exercises in Quantum Mechanics I: The Fundamentals
For A = (Bz, 0,0) and φ = Ez we have
2.4 What is the Schrödinger equation of a system of two particles of masses
and
carrying charges
The Schrödinger equation i
becomes
respectively with
2.5 Starting from the Schrödinger equation for ψ, obtain the equation for ψ∗ .
The Schrödinger equation for ψ is
Taking complex conjugate of the above equation we get
This is the Schrödinger equation for
2.6 If E1 and E2 are the eigenvalues and φ1 and φ2 are the eigenfunctions of a Hamiltonian operator then find whether the energy corresponding to the superposition state
We have
1 + φ2 is equal to E1 + E2 or not
K24365_SM_Cover.indd 22 13/11/14 6:56 PM 1 2 2 2 2 ∂ψ 2
i ∂ψ = 1 2∇ ψ + 2i qBzψ q2 + B z ψ qEzψ . ∂t 2m 2 2 2 c z c2
m1
m2
q1 and q2
1 H = 2m1 p2 + 1 2m2 p2 + q1q2 r12 and r12 = |r1 r2|?
∂ψ = Hψ
∂t
i ∂ψ = ∂t 2 2m1 ∇1ψ 2 2m2 ∇2ψ + q1q2 ψ , r12 2 ∂2 ∂2 ∂2 ∇i = ∂x2 + ∂y2 + ∂z2 , i = 1,2 . i i i
where
2 i ∂ψ = ∂t 2m∇ ψ + V (X,t)ψ .
∗ i = ∂t 2 2m∇ ψ ∗ + V ∗(X, t)ψ ∗
∗
ψ
φ
Hφ1 = E1φ1
Hφ2 = E2φ2
H (φ1 + φ2) = Hφ1 + Hφ2 = E1φ1+ E2φ2 = (E1+ E2) (φ1 + φ2) .
and
Then
Schrödinger Equation and Wave Function 9
2
2.7 Express the Schrödinger equation 2m ∇ + V (x, y,z) ψ(x,y,z) =
Eψ(x, y,z) in the spherical polar coordinates defined by x = rsin θcosφ, y = rsin θsin φ, z = rcosθ and in the parabolic coordinates defined by ξ = r(1 cosθ), η = r(1+ cosθ), φ = φcosθ.
In spherical polar coordinates the Schrödinger equation is
In the parabolic coordinates the Schrödinger equation is
2.8 Find the condition under which both ψ(X, t) and ψ∗(X, t) will be the solutions of the same time-dependent Schrödinger equation.
The Schrödinger equations for ψ(X, t) and ψ∗(X, t) are given by
respectively. The above two equations are identical if V (X, t) = V ∗(X, t), that is if V is real and an even function of t
2.9 What is the major difference between real and complex wave functions?
The probability current density J is given by
If ψ is real then J = 0. If ψ is complex then J need not be zero.
2 10 What is the difference between the wave function ψ1 = ei(kx ωt) and ψ2 = ei(k X ωt)?
ψ1 is a plane (probability)wave travelling along a one-dimensional line. ψ2 is a three-dimensional (probability)wave
K24365_SM_Cover.indd 23 13/11/14 6:56 PM 2
2
2 1 ∂ 2 ∂ 1 ∂ ∂ 2m r2
r + ∂r r2 sin θ ∂θ 1 ∂2 sin θ ∂θ + r2 sin2 θ ∂φ2 + V (r,θ,φ) ψ(r,θ,φ) = Eψ(r,θ,φ)
∂r
2 4 ∂ ∂ ∂ ∂ 1 ∂2 2m ξ + η ξ + η ∂ξ ∂ξ ∂η ∂η + ξη ∂φ2 +V (ξ, η,φ) ψ = Eψ .
2 i ∂ψ ∂t i ∂ψ∗ = 2m ∇ ψ + V (X,t)ψ , 2 = 2 ∂t 2m ∇ ψ ∗ + V ∗(X, t)ψ
J = 2mi [ψ ∗∇ψ ψ∇ψ ∗]
10 Solutions to the Exercises in Quantum Mechanics I: The Fundamentals
2 11 Write the Hamiltonian of a photon
For a photon H = E = mc2 = mcc = pc.
2.12 What is the physical meaning of �x� = 0?
ψ is symmetric or antisymmetric with respect to x = 0
2 13 Write the operatorforms of kinetic energy and angular momentum L = r × p.
The operator form of kinetic energy mv2/2 is
For L = r × p we have with
The components of L are
2.14 Consider the one-dimensional Schrödinger equation u(x) = dx ln ψ(x). Obtain the Schrödinger equation under the change of variable d u(x) = dx ln ψ(x)
The general transformation is u = ψx/ψ or ψx = −uψ Then ψxx = −uxψ uψx.Now, the Schrödinger equation is rewritten as (with λ = 2mE/ 2 and g = 2mV/ 2)
That is, ux u2 λ + g = 0
K24365_SM_Cover.indd 24 13/11/14 6:56 PM − ∇
1 2 p2 2 2 (K.E.)op = 2 mv = = 2m 2m
= ix
i j k L = r × p = x y z px py pz = i (ypz zpy) +j (zpx xpz) + k (xpy ypx)
∂ ∂ ∂ ∂ Lx = i y∂z + z ∂y = i z y , ∂y ∂z ∂ ∂ ∂ ∂ Ly = i z∂x + x∂z = i x z , ∂z ∂x ∂ ∂ ∂ ∂ Lz = i x ∂y + y∂x = i y x ∂x ∂y d
r
+ jy+ kz
−uxψ u
ψ
uxψ+ u2ψ + (λ g)ψ = 0
ψx+ (λ g)
= 0 or
.
Schrödinger Equation and Wave Function 11
2 15 Find the conditions to be satisfied by the functions f and g such that under the transformation ψ = f(x)F(g(x)) the Schrödinger equation ψ
+ (E V )ψ = 0 can be written as
In terms of f and F the Schrödinger equation is written as
= 0 , (2.3) where prime denotes differentiation with respect to x Since F is F (g(x)) we have
Then
Comparison of Eqs (2 1) and (2 5) gives
The conditions on f and g are given by Eqs. (2.6). From Eqs. (2.6) we obtain
(2 7a) we find f /f and substituting it in (2 7b) we obtain
Substituting the above in Eq. (2.7b) we obtain the Eq. (2.2).
K24365_SM_Cover.indd 25 13/11/14 6:56 PM
xx
Fgg + Q(g)Fg + R(g)F
1)
that E V = (g )2 R 1 Q Q2 g − + g 2 (2 2) 2 g 4 2g 2g
(g) = 0 (2
Then show
fF
F
f F
V
F
F = Fgg , F = Fgg(g )2 + Fgg . (2.4)
Eq (2 3)
2f g f (E V ) F = 0 . (2.5) Fgg + + g f g 2 Fg + g 2f 2 + g 2
2f Q = + g f g g 2 , R = f g 2f 2 + E V g 2 . (2.6)
+ 2f
+
+ (E
)f
becomes
f g Q = f 2 g 2g
E V = g 2R f
f
That is, d f = dx f f f f 2 f2 . (2.8) f = f = d f + dx f d g Q dx 2 f 2 f2 g 2g g Q + 2 g 2 2g 1 = 2Qg g + 2g g 2Q2 4 g 2 + 2g (2 9)
(2 7a)
(2 7b)
From
12 Solutions to the Exercises in Quantum Mechanics I: The Fundamentals
2 16 What are the effects of addition of a constant to a potential on the time-independent Schrödinger equation and the energy levels?
The time-independent Schrödinger equation with the addition of a constant α to a potential is given by
This can be rewritten as
where E� = En α. Therefore ψn remains the same but the energy eigenvalues of the new system are En α.
2 17 Which of the following wave functions are admissible in quantum mechanics? State the reasons
The conditions to be satisfied by a wave function are: (i) ψ should be normalizable. (ii) It should be single-valued. (iii) It must be finite at every point. (iv) It and its first partial derivatives must be continuous.
The functions (a), (b), (e), (j) and (k) satisfy all the above conditions and hence they are admissible wave functions (verify).
The functions (c), (d), (f), (h), (i) are nonnormalizable. The function (g) is multivalued. Therefore, they are not admissible wave functions
2.18 Normalize
2 19 Find the value of N for which the wave function ψ(x)
and 0 for |x| > a is normalized. The normalization condition gives
K24365_SM_Cover.indd 26 13/11/14 6:56 PM n n 2 2 a 2 √ ∞ 1 = ∞ 2 a a
2 2 2m ∇
+ V + α ψn = Enψn 2 2 2m ∇ + V ψn = (En α) ψn = E� ψn ,
(a) e x . (b) sechx. (c) e x . (d) tanhx. (e) sin x, 0 < x < 2π. (f) sin x, ∞ < x < ∞ . (g) 1 < x < 1. e x2 (h) tan x (i) secx (j) xe x (k) 1 x2 ,
2 2
= Neikx x /(2σ ) . From the normalization condition we obtain ψ ∗ ψ dx = N2σ −∞ −∞ e y dy = N2σ √ π
= 1/(σ √ π )1/2
the wave function ψ
Thus N
N
a
1 = ψ ∗ ψ dx = N2 dx = N2x|a
N = 1/ √ −a a = 2aN2 a
=
for |x| <
Thus,
Schrödinger Equation and Wave Function 13
2 20 Normalize the wave function ψ =
|x| sin αx It is given that
condition gives
2.21 A particle of mass m moves in a one-dimensional box of length L with origin as the centre. If the wave function of this particle is ψ(x) =
) for |x| < L/2 and 0 otherwise find the factor
N can be determined from normalization condition We obtain
If L is very small then N
K24365_SM_Cover.indd 27 13/11/14 6:56 PM 0 0 2 ∞ 3 5 7
∞ e sin αx dx = α /(1 + α ) −x 2 2 0
normalization
1 = N2 ∞ e 2|x| sin2 αx dx −∞ 0 ∞ = N2 −∞ e2x sin2 αxdx + N2 0 e 2x sin2 αx dx = 2N2 ∞ e 2x sin2 αx dx 0 N2 = e 2x 2 |∞ N2 ∞ 2 0 e x cosαx dx N2 = 2 N2 2 e x cosαx|∞ α 0 e x sin αx dx N2 = 2 N2 1 2 α3 1 + α2 N2α3 = 2(1 + α2) That is, N = 2(1 + α2)/α3
e
The
2
Nx(1 x
N
1 = N2 L/2 x2 1 x2 dx L/2 L/2 = N2 L/2 x2 2x4 + x6 dx = N2 L L + L 12 160 448 Thus, L3 N = 12 L5 + 160 L7 1/2 . 448
3
≈ 12/L
14 Solutions to the Exercises in Quantum Mechanics I: The Fundamentals
2 22 A quantum mechanical particle moving in one-dimension has the wave function
where
and b are constants (b > 0). Find the probability that the position of the particle lies in the region ∞ < x ≤ b
First, we normalize the wave function. We obtain
K24365_SM_Cover.indd 28 13/11/14 6:56 PM b b
ψ(x) = cxe |x|/b , ∞ < x < ∞
c
1 = ∞ ψ ∗ ψ dx −∞ = c2 ∞ x 2e −∞ 0 2|x|/b dx ∞ = c2 −∞ x2e2x/b dx + c2 0 x2e 2x/b dx = 2c2 ∞ x 2e 0 2x/b dx Substituting 2x/b = y in the integral, we get b3c2 ∞ 1 = 4 Integrating by parts we get b3c2 ∞ y2e y dy 0 3 2 ∞ 3 2 1 = ye y dy = b c 2 0 2 0 e y dy = b c . 2 That is c2 = 2/b3 or c = 2/b3 The probability P ( ∞ < x ≤ b) is P = ψ ∗ ψ dx −∞ 2 0 = b3 x2e2|x|/b dx + b x2e 2x/b dx −∞ 2 b3 0 = b3 8 ∞ 1 ∞ 0 y2e y dy + 0 b y2e y dy = y 2e y dy + 4 0 0 y2e y dy Integrating by parts we get P = 1 e b b2 4 + 2b + 2
Schrödinger Equation and Wave Function 15
2 23 What is the probability current density corresponding to ψ(x) = Ae αx where A and α are constants?
The probability current density is given by
Since the given ψ is real, ψ∗ = ψ,
2 24 A free particle in one-dimension is in a state described by ψ = A ei(pxx Et)/ + B e i(pxx+Et)/ where px and E are constants Find the probability current density J is given by
2.25 If the wave function of a particle at t = 0 is ψ(x, 0) = Ne ikx (x /2a ) calculate the probability density and current density
First, we normalize the given wave function. We obtain
The probability density is obtained as
Then the current density is determined as
K24365_SM_Cover.indd 29 13/11/14 6:56 PM 2 2 ψ
J =
ψ
2mi [ψ ∗∇ψ
∇ψ ∗]
J = 2mi [ψ∇ψ ψ∇ψ]
= 0 .
J = = m Im(ψ ∗∇ψ) Im A∗ e i(pxx Et)/ + B∗ ei(pxx+Et)/ m ipx + Be × px 2 Aei(pxx Et)/ 2 i(pxx+Et)/ = m |A| |B| 2 2
1 = ∞ ψ ∗ ψ dx = 2N2a ∞ e y 2 dy = N2a √ π −∞ 0 That is N = (1/(a2π))1/4
P (x) = 1 a √ π e x /a .
J(x) = 2mi k dψ ψ ∗ dx 1 2 2 dψ∗ dx = m a √ π k e x /a = P (x) . m
16 Solutions to the Exercises in Quantum Mechanics I: The Fundamentals
2 26 For the Gaussian wave function ψ(x) = the probability current density
The probability current density is given by For any real ψ
Then J = 0 Therefore, for the real Gaussian function J = 0
2 27 Verify whether the wave function ψ = Neikx x /(2a ) satisfies the continuity equation or not.
The continuity equation is given by ∂ρ dJ +
dx = 0 For the given function
Now the continuity equation becomes xe x /a = 0 which is true only if
x = 0 Hence, the given ψ does not satisfy the continuity equation
2 28 The wave function of a linear harmonic oscillator with potential V = mω2x2/2 is ψ = xe mωx /(2 ) Find its energy
The Schrödinger equation for a potential V is
For the given potential and ψ the above equation becomes
From the above equation we get E =
K24365_SM_Cover.indd 30 13/11/14 6:56 PM 2 2 2 2 2 2 2
1 σ √ π 1/2 e x /2σ
calculate
J = 2mi dψ ψ ∗ dx dψ∗ ψ dx . dψ dψ∗ ψ ∗ dx ψ dx = 0
2 2
ρ = ψ∗ψ = N2e x /a and ∂ρ = 0 Further ∂t J = Im ψ ∗ m dψ dx 2 2 = Im N2e x /a m ik x/a2 Then N2 k = m dJ e x /a N2 k 2 2 dx = 2 ma2 xe x /a 2 2
∂t
2m ψxx + 2 (E V )ψ = 0
3mω m2ω2 3 2mE m2ω2 3 x + 2 x + 2 x 2 x = 0
ω
3
/2
Schrödinger Equation and Wave Function 17
2 29 If ψ = 1/Lcos(πx/(2L)) for the system confined to the potential V (x) = 0 for |x| < L and ∞ for |x| > L then calculate E
The Schrödinger equation for the given system is
2 30 The wave function of a particle confined to a box of length L is 2/L sin(πx/L) in the region 0 < x < L and zero everywhere else. Calculate the probability of finding the particle in the region 0 < x ≤ L/2
2 31 Write the law of conservation of energy H = T + V in terms of expectation values We write
2 32 Are the wave
of the electron in hydrogen atom orthogonal?
0, the given two wave functions are orthogonal
K24365_SM_Cover.indd 31 13/11/14 6:56 PM L 0 0 ∗ 2 a a ∞ ∞
2m
ψxx + 2 Eψ = 0 . 2 2 π 2 πx 2π2 Eψ = 2mψxx = 2m √ 2L cos = ψ . 2L 8mL2
E = 2π2/(8mL2
Then
Thus.
).
obtain P (0 < x ≤ L/2) = L/2 0 ψ ∗ ψ dx 2 L/2 = sin2 πx dx L 0 L 1 L/2 = 1 cos 2πx 1 dx = L 0 L 2
We
�H�
�T� + �V �.
=
1 1/2 r/a0 1 1/2 r r/(2a0) ψ1 = πa3 e and ψ2 = 32πa3 2 − e 0
The orthogonality integral is calculated as ∞ ψ ψ dτ ∞ r r e 3r/2a0 dr 1 2 ∝ 2 0 0 0 ∞ ∞ ∝ 6 y2e y dy 2 0 0 ∝ 6 y2e y dy 6 0 0 y3e y dy y2e y dy = 0 . Since ∞ ψ ∗ ψ dτ =
1 2 0
functions
18 Solutions to the Exercises in Quantum Mechanics I: The Fundamentals
2 33 Find the potential corresponding to the following wave functions:
(a)
(b)
Then
(b) For the given wave function we obtain
(c) Using the given wave function we get
K24365_SM_Cover.indd 32 13/11/14 6:56 PM 2 −
ψ
1 π1/4 e x /2 , E = ω/2
=
ψ
sin(π
< L
0
> L, E
2π2/(8mL2)
ψ = αxe βxeiE/ (d) ψ = (mV0/ )e kx for x > 0, (mV0/ )ekx for x < 0 and E = mV0/(2 2), k2 = mV0/ 2 .
= A
x/(2L)) for |x|
and
for |x|
=
(c)
ψ = Eψ
2 d2ψ 2 2m dx2 + V ψ= Eψ or V ψ= 2mψxx+ Eψ
the
ψ and E in
above equation we get 2 V ψ= d ( x)ψ+ Eψ = 2 ψ + x 2 ψ + Eψ
(a) The eigenvalue equation H
is
Substituting
given
the
2m dx 2 2m 2 ω V = 1 + x2 + E = 2m x2 1 + . 2m 2
V ψ = 2 d π πx Acos + Eψ 2m dx 2L 2L 2π2 = 8mL2ψ + Eψ = Eψ+ Eψ = 0
< L
∞
|x| > L.
2 V ψ = αe iE/ d 2m dx 2 e βx βx2e βx + Eψ = αeiE/ βe βx 2βxe βx + β2 x 2e βx + Eψ 2m or 2 V αxe βxeiE/ = αeiE/ βe βx 2βxe βx + β2 x 2e βx + Eψ 2m
Therefore, V (x) = 0 for |x|
and
for
Schrödinger Equation and Wave Function 19
2.34 A particle of mass m is confined in the infinite square-well potential V = 0 for 0 < x ≤ L and V = ∞ otherwise. It has the normalized stationary state eigenfunctions
(x) and eigenvalues
K24365_SM_Cover.indd 33 13/11/14 6:56 PM φ ∗ 2 e
is, 2 β V = 2m x 2β + β2x + E
That
x > 0 2 V ψ= d ( kψ) + Eψ = 2 k2 ψ + Eψ 2m dx 2m
V = k2 2 1 . Similarly, for x < 0 we obtain V = 2 m k2 2 2 m 1
(d) For the given wave function for
Then
φ
En = n2 2π2/(2mL2)
wave function
t
by ψ(x, 0) = (φ1(x) +φ2(x))/ √ 2 What is the
time τ for which ψ(x, t) will be orthogonal to ψ(x, 0)? The wave function ψ(x, t) is 1 ψ(x, t) = √ 2 φ1e iE1t/ + φ2e iE2t/ At t = τ the condition for ψ(x, 0) and ψ(x, t) to be orthogonal is L We get L ψ ∗(x, τ)ψ(x, 0) dx = 0 . 0 iE1τ/ 1 0 + φ ∗ eiE2τ/ (φ1 + φ2) dx = 0 Since L φ φ dx = δ we get eiE1τ/ + eiE2τ/ = 0 That is 0 m n E1τ mn E1τ E2τ E2τ cos + isin + cos + isin = 0 . Equating the real and imaginary parts to zero separately we get cos E1τ = cos E2τ , sin E1τ = sin E2τ We rewrite the above conditions as cos E1τ = cos π + E2τ , sin E1τ = sinπ + E2τ . Thus, τ = π/(E1 E2). Since E2 > E1 we write τ = π/(E2 E1).
n
Its
at time
= 0 is given
smallest positive
20 Solutions to the Exercises in Quantum Mechanics I: The Fundamentals
2.35 Show that �xpx + pxx� is real.
2 36 Show that
2 37 Calculate
K24365_SM_Cover.indd 34 13/11/14 6:56 PM � x x 2 x 2 2 † x ∞ 0
�pxx�
∞ �pxx� = −∞ ψ ∗ pxxψ dx = ∞ (p −∞ ∞ ψ)∗ xψ dx ∗ = −∞ = ∞ −∞ pxψx ∗ ψ ∗ dx ∗ ψ ∗ xpxψ dx = �xpx�∗ Now, �xpx + pxx� = �xpx� + �pxx� = �xpx� + �xpx�∗ = 2Re�xpx� Thus �xpx + pxx� is real
Consider
We obtain
�
� = �A n
Let Aψ = αψ
�An� = ∞ ψ ∗Anψ dτ −∞ = ∞ ψ A Aψ dτ ∗ n 1 −∞ = α ∞ ψ ∗An 1 ψ dτ −∞ = αn ∞ ψ ∗ ψ dτ −∞ We find �A� = n = αn ∞ ψ ∗Aψ dτ = α ∞ −∞ −∞ ψ ∗ ψ dτ = α. Hence, �An� = αn = �A� .
An
in its eigenstates.
Then we obtain
2�
ψ
√ e k|x| x k We obtain �p2� = ψ ∗ p2ψ dx = 2 ∞ e k|x| d dx2 e k|x| dx −∞ −∞ The
rewritten as �p2� = 2k ekx d dx2 ekx dx + ∞ e kx d dx2 e kx dx −∞ 0
�p
for
=
above equation can be
Then we obtain
2 38 A particle of mass m in the one-dimensional energy well V (x) = 0 for 0 ≤ x ≤ L and ∞ otherwise is in a state whose wave function is given by ψ(x) = Nx(L x) where N is the normalization constant Determine �E� in this state.
The normalization condition gives
is,
2 39 Show that
K24365_SM_Cover.indd 35 13/11/14 6:56 PM x 2 ∞ 2 x 0 L � �
Schrödinger Equation and Wave Function 21
�p2� = 2k3 e2kx dx + ∞ e 2kx dx = 2k2 −∞ 0
L 2 5 1 = N2 x2(L x)2 dx = N L 0
30/L5 . Next, L �E� = 0 30 ψ ∗Hψ dx 2 d2 = ψ 0 2m dx2 2 L ψ dx = N2 m 0 xL x2 dx 5 2 = mL2 d 2 i
x2 = dt m �xpx� m �x2� is given by �x2� = ∞ ψ ∗ x2ψ dx Then −∞ d 2 d ∞ 2 dt �x � = ψ ∗ x ψ dx dt −∞ = ∞ ∂ψ∗ ∂ψ x ψ dx + ψ ∗ x dx −∞ ∂t ∂ψ −∞ ∂t ∂ψ∗
Schrödinger equation for and ∂t ∂t we get d 2 1 ∞ p2 ψ ∗ ∗ 2 dt �x � = i −∞ 1 ∞ + V ψ 2m p2 x ψdx + i ψ ∗ x2 −∞ x ψ + V ψ dx 2m i = 2m
That
N =
Using the
K24365_SM_Cover.indd 35 13/11/14 6:56 PM x x ∞ ψ ∗ p 2 −∞ x2ψ i dx − 2m ∞ ψ ∗ x 2p2ψ dx . −∞
22 Solutions to the Exercises in Quantum Mechanics I: The Fundamentals
K24365_SM_Cover.indd 36 13/11/14 6:56 PM x 2 � �
the first integral we obtain d 2 i ∞ 2 2 2 dt �x � = 2m i ψ ∗ 2pψpxx −∞ ∞ + ψp x dx = 2m 2 ψ ∗ 4i xpxψ+ 2ψ( i )2 dx −∞ i = m �xpx� m . 2 40 Show that d �x2� = 2 �p2� + 2 �xF�. dt2 d m2 x m 2 i We have x 2 = dt m �xpx� m Then d2 2 2 d dt2 �x � = m dt �xpx� = 2 d ∞ ψ ∗(xpx)ψ dx m dt −∞ = 2 ∞ ∂ψ∗ xpxψ dx + 2 ∞ ∂ψ ψ ∗ xpx dx m −∞ ∂t m −∞ ∂t
the Schrödinger equation d2 2 i ∞ 2 2i ∞ dt2 �x � = m2 pxψ ∗ xpxψ dx + m −∞ V ψ ∗ xpxψ dx −∞ i ∞ 3 2i ∞ m2 ψ ∗ xpxψ dx m −∞ ψ ∗ xpx(V ψ) dx −∞ = i ∞ ψ2 px(pxxpxψ+ xp2 ψ) xp3 ψ dx m2 x x −∞ 2i ∞ dV m ψ ∗ xψ −∞ −i dx dx Substituting F = dV/dx we get d2 2 i ∞ 2 2 dt2 �x � = m2 2i ψ ∗ 2pxxpxψ dx + m �xF � −∞ ∞ 2 = m2 2 ψ ∗( i )p2 ψ dx + −∞ 2 m �xF�
Expanding
Using
K24365_SM_Cover.indd 36 13/11/14 6:56 PM x = m2 �p2� + m �xF�
K24365_SM_Cover.indd 37 13/11/14 6:56 PM x x x x x x i i x x d dt x m Schrödinger Equation and Wave Function 23
Show that �p2 � = �2F px + pxF�. By definition �p2 � = ∞ ψ ∗ p2 ψdτ. Then −∞ d 2 ∞ ∂ψ∗ 2 ∞ 2 ∂ψ dt�px� = pxψ dτ + ∂t −∞ ψ ∗ px ∂t dτ −∞
2.41
obtain d 2 1 ∞ p2 ψ ∗ ∗ 2 dt�px� = i −∞ 1 ∞ + V ψ 2m p2 ψ pxψ dτ That is, + i ψ ∗ p2 −∞ x + V ψ 2m dτ d 2 i ∞ 2 i ∞ 2 dt�px� = V ψ ∗ pxψ dτ −∞ ψ ∗ px (V ψ) dτ −∞ i ∞ ∞ = 2 ψ ∗ pxV pxψ dτ + −∞ ψ ∗ p2 V ψdτ −∞ ∞ = 2 ( i ) −∞ ∞ ψ ∗∇V pxψ dτ ( i ) −∞ ψ ∗(px∇V )ψ dτ = 2 ∞ ψ ∗Fp ψ dτ + ∞ ψ ∗ p F ψdτ −∞ −∞ = �2F px + pxF � 2 42 Show that d dt�pxx� = 1 �p2� + �xF � The expectation value of pxx is ∞ �pxx� = ψ ∗ pxxψ dx −∞ ∞ ∞ = i −∞ ψ ∗ ψ dx + −∞ ∞ ψ ∗ xpxψ dx Now = i + −∞ ψ ∗ xpxψ dx d dt�pxx� =
Using the Schrödinger equation we
K24365_SM_Cover.indd 37 13/11/14 6:56 PM x d ∞ dt −∞ ψ ∗ xpx ψ dx = ∞ −∞ ∂ψ∗ xpx ∂t ψ dx + ∞ ψ ∗ xp −∞ ∂ψ dx . ∂t
24 Solutions to the Exercises in Quantum Mechanics I: The Fundamentals We
K24365_SM_Cover.indd 38 13/11/14 6:56 PM px ∂ψ x ∗ ∗ x x x x x x x ∞ ∗ ∞ m � � m
have 2 ∗ 2 i ∂ψ = ψ + V ψ , i p = ψ + V ψ ∂t 2m ∂t 2m
these two equations for ∂ψ/∂t and ∂ψ∗/∂t we get d i ∞ 2 ∗ i ∞ dt �pxx� = 2m pxψ xpxψ dx + −∞ V ψ ∗ xpxψ dx −∞ i ∞ 3 i ∞ 2m ψ ∗ xpxψ dx −∞ ψ ∗ xpx(V ψ) dx −∞ i = 2m ψ ∗ p2 (xpxψ)dx −∞ i 2m ∞ ψ ∗ xp3 ψ dx −∞ Then + ∞ ψ −∞ xψF dx . d i ∞ 2 i ∞ 3 dt �pxx� = = 2m i 2m ψ ∗ px(xpxψ)dx 2m −∞ ψ ∗ pxxp2 ψ dx + �xF� −∞ ψ ∗ xpxψ dx + �xF� −∞ = 1 ∞ ψ ∗ p2 ψ dx + �xF� m −∞ = 1 �p2� + �xF� 2 43 For a free particle show that d p2 = 0 and dt d dt �xpx + pxx� = 2 �p2�
result d 1 2 d 2 dt �pxx� = m �px� + �xF� , For a free particle F = 0 and hence dt �px� = �2Fpx + pxF� d 1 2 d 2 dt �pxx� = m �px� , dt �px� = 0 Consider [x,px] = xpx − pxx= i From this we write xpx pxx+ pxx pxx = i xpx + pxx 2pxx = i xpx+ pxx = i + 2pxx
Using
We have the
K24365_SM_Cover.indd 38 13/11/14 6:56 PM = 2 Then d d 2 2 dt �xpx + pxx� dt �pxx� = m �px� .
2 44 φ1 and φ2 are the only eigenfunctions of a system belonging to the energy eigenvalues E0 and E0 respectively In a measurement of the energy of the system, �E� is found to be E0/2 Find the wave function of the system
The
2 45 The H of a charged particle in uniform electric and magnetic fields
Both electric field E and magnetic field B are applied along the zdirection Applying Ehrenfest’s theorem, show that
K24365_SM_Cover.indd 39 13/11/14 6:56 PM = 2m
Schrödinger Equation and Wave Function 25
is ψ = C1φ1 + C2φ2 , C2 + C2 = 1 Further 1 2 �E� = C2E1+ C2E2 1 2 2 2 = C1 E0 C2 E0 = E0 C2 C2 1 2 Since �E� = E0/2 we get C2 C2 = 1/2. Solving C2 + C2 = 1 and C2 C2 = 1/2 we get C √ 1 = 3/2 2 and C = 1/2 1 2 Then 1 − 2 1 √ 3 ψ = 2 φ1 + 2 1 2φ2 .
wave function of the system
1
H = (p − eA)2 − eE0k.k where A = B0( y,x, 0)/2.
is given by
d dt�r� = 1 m �p eA� and d dt�p eA� = eE0k + e m �p eA� × ∇ ×A 1 2
× A
× A = B0k H is d (p eA) 2m eE0 Now,
dt�x�
d dt�x� = ∞ ∂ψ∗ xψ dx + −∞ ∂t ∞ ∂ψ ψ ∗ x dx −∞ ∂t 1 ∞ i −∞ Hψ ∗ xψ dx + 1 ∞ i −∞ ψ ∗xHψ dx 1 = i�[x,H]� .
∇
is obtained as ∇
consider
. We obtain
26 Solutions to the Exercises in Quantum Mechanics I: The Fundamentals
Next, we find
2 46
Hence,
Hence,
A particle of mass m enclosed in a one-dimensional box of length L such that 0 < x < L, has energy eigenfunctions
)
t
where B is a constant.
0, the particle has the wave function
(i) If the energy of the particle is measured at t = 0, what are the
K24365_SM_Cover.indd 40 13/11/14 6:56 PM x x m m m 1 1
[x,H] = 1 x, 2m(p eA) · (p eA) +eE0 1 1 = x, p2 + 2m 1 x, e2A2 2m 1 2m[x,p · eA] 2m[x,eA · p] + [x,eE0] = 1 x, p2 2m i i 2m (i eAx) (i eAx) 2m
= m px m eAx d 1 i eAx 1 dt�x� = i px i m = (px eAx)
Therefore,
d 1
dt�r� = m �p eA� d2 d d d dt2 �r� = �F� = dt m dt�r� = dt�p eA� 1 = i�[p eA,H]�
Next,
e �F� = eE
v × B − B × v� = eE +
2m
We obtain with v = dr/dt, B = ∇ × A
+ 2m�
e
[(p eA) × B B × (p eA)]
= eE + e [(p eA)
∇ × A] d e dt�p eA� = �F� = eE0k + m (p eA) × ∇ × A
×
n = 1, 2,··· and En = n2π2 2
2
ψ
φn(x) = A sin(nπx/L),
/(2mL
At
=
(x) = B sin(2πx/L) cos(πx/L)
Schrödinger Equation and Wave Function 27
possible results of the measurement? (ii) What is the expectation value of energy?
(i) The wave function can be written as a combination of eigenfunctions as
That is, the system can have only two eigenstates φ1 and φ3 each with probability B2/4. Further, (B2/4) + (B2/4) = 1 gives B =
Therefore,
If the energy is measured then one may get the energy as E1 with the probability 1/2 and E3 with probability 1/2.
(ii) We obtain
2 47 Given the normalized ground state wave function of hydrogen atom
find the expectation value of its z-coordinate.
K24365_SM_Cover.indd 41 13/11/14 6:56 PM 2 0 0 0
ψ = B sin 2πx L πx cos L B πx = sin 2 L B + sin 3πx L = 2 (φ1 + φ3)
√
1 ψ = √ 2 (φ1 + φ3)
�E� = C2E1+ C2E3 = 1 (E1 + E3) = 5π2 2 1 3 2 2mL2
ψ100 = 1/(πa3)1/2 e
0
We obtain �z� = ∞ π 2π 1 πa2 e 2r/a0 zr2 dr sinθdθdφ . 0 0 0 0 Substituting z = rcosθ we get �z� = 1 πa3 ∞ π 2π r3e 2r/a0 sin θcosθdr dθ dφ 0 0 0 1 0 π ∞ 3 2r/a = 2a3 cos2θ|0 r e 0 dr = 0
r/a
28 Solutions to the Exercises in Quantum Mechanics I: The Fundamentals
2 48 If Hφn(x) = Enφn(x) find the expectation value of the Hamiltonian operator H in the normalized superposition state ψ(x, t) =
Cnφn(x) e iEn
. We obtain
2 49 Consider a system in a state ψ = (φ1 + φ2)/
where φ1 and φ2 are orthonormal eigenfunctions with the eigenvalues E1 and E2 respectively. What is the probability of finding the system in the energy E1? What is E ?
The probability of finding the system with energy E1 is (1/ √ = 1/2.
2 50 Consider a spherically symmetric potential energy function given by V (r) = 0, for 0 < r < a and ∞ for r > a. Given the solution ψ(r) = A sin kr r coskr + B r where k = (2mE/ 2)1/2 satisfying the Schrödinger equation, obtain the corresponding eigenvalues by applying proper boundary conditions
As r → 0, ψ must be finite The condition limr→0 ψ(r) = finite sets sin kr B = 0. So ψ(r) = A r At r = a we require ψ = 0. This gives sin ka = 0 That is, ka = nπ, n = 1,2, Hence, the energy eigenvalues are given by
K24365_SM_Cover.indd 42 13/11/14 6:56 PM = C∗ n n φ ∗ = CnC e E 2 2)2
∞ n=0
E = ψ|H|ψ n Cnei(En En)t/ ∞ φ ∗ Hφn dx n n −∞ ∞ = CnC∗ ei(En En)t/ n Enφn dx n n −∞ ∗ i(En En)t/ n nδnn n n = |Cn|2En . n
t/
√
Next 2 E = |Cn|2En = n=1 1 2 E1 + 1 2 E2 = 1 2 (E1 + E2)
En = 2k2 n = 2m n2 2π2 2ma2
